$f(x, y, z) = \left( xe^y, \cos(xz), x + z \right)$ $\text{div}(f) = $
Answer: The formula for divergence in three dimensions is $\text{div}(f) = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z}$, where $P$ is the $x$ -component of $f$, $Q$ is the $y$ -component, and $R$ is the $z$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial P}{\partial x} &= \dfrac{\partial}{\partial x} \left[ xe^y \right] \\ \\ &= e^y \\ \\ \dfrac{\partial Q}{\partial y} &= \dfrac{\partial}{\partial y} \left[ \cos(xz) \right] \\ \\ &= 0 \\ \\ \dfrac{\partial R}{\partial z} &= \dfrac{\partial}{\partial z} \left[ x + z \right] \\ \\ &= 1 \end{aligned}$ Adding the three partial derivatives, $\text{div}(f) = e^y + 1$.